Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

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Question:

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

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Answer: Octane will be used completely.Explanation:To calculate the moles :[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    [tex]\text{Moles of octane}=\frac{2.3g}{114g/mol}=0.0202moles[/tex][tex]\text{Moles of oxygen}=\frac{12.4g}{32g/mol}=0.388moles[/tex]The balanced chemical reaction will be[tex]2C_8H_{18}+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]  According to stoichiometry :2 moles of octane require = 25 moles of [tex]O_2[/tex]Thus 0.0202 moles of octane will require=[tex]\frac{25}{2}\times 0.0202=0.2525moles[/tex]  of [tex]O_2[/tex]Thus octane is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.  Thus octane will be used completely.

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